The comments state the objective is to print to 2 decimal places.
There's a simple answer for Python 3:
>>> num=3.65
>>> "The number is {:.2f}".format(num)
'The number is 3.65'
or equivalently with f-strings (Python 3.6+):
>>> num = 3.65
>>> f"The number is {num:.2f}"
'The number is 3.65'
As always, the float value is an approximation:
>>> "{}".format(num)
'3.65'
>>> "{:.10f}".format(num)
'3.6500000000'
>>> "{:.20f}".format(num)
'3.64999999999999991118'
I think most use cases will want to work with floats and then only print to a specific precision.
Those that want the numbers themselves to be stored to exactly 2 decimal digits of precision, I suggest use the decimal type. More reading on floating point precision for those that are interested.
Answer from Andrew E on Stack OverflowThe comments state the objective is to print to 2 decimal places.
There's a simple answer for Python 3:
>>> num=3.65
>>> "The number is {:.2f}".format(num)
'The number is 3.65'
or equivalently with f-strings (Python 3.6+):
>>> num = 3.65
>>> f"The number is {num:.2f}"
'The number is 3.65'
As always, the float value is an approximation:
>>> "{}".format(num)
'3.65'
>>> "{:.10f}".format(num)
'3.6500000000'
>>> "{:.20f}".format(num)
'3.64999999999999991118'
I think most use cases will want to work with floats and then only print to a specific precision.
Those that want the numbers themselves to be stored to exactly 2 decimal digits of precision, I suggest use the decimal type. More reading on floating point precision for those that are interested.
The simple way to do this is by using the round buit-in.
round(2.6463636263,2) would be displayed as 2.65.
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You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.
With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).
Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.
For example,
>>> 125650429603636838/(2**53)
13.949999999999999
>>> 234042163/(2**24)
13.949999988079071
>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999
If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:
- Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
- Or use a fixed point number like decimal.
There are new format specifications, String Format Specification Mini-Language:
You can do the same as:
"{:.2f}".format(13.949999999999999)
Note 1: the above returns a string. In order to get as float, simply wrap with float(...):
float("{:.2f}".format(13.949999999999999))
Note 2: wrapping with float() doesn't change anything:
>>> x = 13.949999999999999999
>>> x
13.95
>>> g = float("{:.2f}".format(x))
>>> g
13.95
>>> x == g
True
>>> h = round(x, 2)
>>> h
13.95
>>> x == h
True
I'm not sure if this is a problem with my IDE or if its a Python issue but when I add a floating point number, take 3.14 for example, to some numbers I get a lot of additional 0s.
y = 3.14 y1 = 3.14 + 1 print(3.14) Output: 4.140000000000001
I noticed this issue occurs when I add 1 to 4 but if I add 5 I get 8.14 instead, without all of the 0s.
Why does this only happen when adding certain numbers?
When the matter is precision - like, really precise - floats are crap. Even double precision can be a gamble when working with scientific data - very small and very large numbers.
What are the best options when I need to work with numbers on the -15th and +20th orders of magnitude? (at the same time, mind you)
Is the decimal.py module precise enough for those sorts of calculations? If not, is it possible to get precise results with python or will I have to write some matlab modules to crunch my numbers?
Matlab uses floats too. Why would writing matlab code solve floating point issues that you get in python?
Have you checked what kind of floating point noise your application can tolerate? This isn't a python issue. It's going to apply so long as you choose to use a computer.
NumPy defines a range of data types, maybe one of those is what you want?
I know this is not a Python-specific issue. I read some posts saying it's best to cast the float as a decimal, and others say round before/after calculation. What are your recommendations? Is there a "pythonic" way to do this?
Edit: Decimal